>(2) I am actually using the "earth is a perfect sphere" for the 
>calulations at the moment as I haven't yet come across anything 
>giving me the maths behind the oblate spheroid. I know that there is 
>two different latitudes involved when calculating or dealing this, I 
>think it has something to do with the real latitude involves 
>calculations with the normal to the surface. All the locations are 
>stored in a database giving their name, latitude, longitude and 
>altitude and all the satellites are stored as their special element 
>files and the passed through the orbit propogation function to 
>obtain their X,Y,Z coorindate that is then also converted to a 
>latitude, longitude and altitude.
>
>Once again thanks for the help,
>
>Regards,
>
>   Ashley ~)~

Ashley ~)~

I'm not sure that using oblate spheroid calculations would do that 
much better in defining the size and configuration of the "footprint" 
than using a perfect sphere in this problem. The reason being is that 
the height of the orbit of the satellite depends upon the pull of 
gravity. Considering that gravity changes depending upon how much 
mass is between the satellite and the center of the earth; then, so 
does the altitude of the satellite, the satellite would follow a 
gravitational gradient like sea level; thus, one (altitude) would 
tend to cancel the other (Earth configuration) to some effect with 
respect to the size of the footprint. In simpler words, the ultimate 
size of the "footprint" may not change appreciatively.

tedd
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